Question

if two
adjacent anglies of a parallelogramm are in the
ratio 4:5, then find the measure
of all angles​

Answer 1

Answer :-

The all angles of parallelogram are :-

80°, 100°, 80° and 100°.

Step-by-step explanation:

Given that,

• The two adjacent angles of parallelogram are in the ratio of 4:5

Assumption:

Let us assume the ratio of adjacent angles of parallelogram as 4x and 5x.

And

Let the Parallelogram be ABCD,

[∠A = 4x] ; [∠B = 5x].

We know,

• The sum of ajacent angles of parallelogram measures 180°

∴ 4x + 5x = 180°

⠀⠀⠀⠀⠀ ⠀____________________

⇒ 4x + 5x = 180

⇒ 9x = 180

⇒ x = 180/9

⇒ x = 20

We got, The value of x as 20.

Now, According the question,

→ The measure of all angles :-

• 4x [∠A] = 4*20 = 80°
• 5x [∠B] = 5*20 = 100°

Also,

→ ∠C = ∠A = 80° ... opposite angles

→ ∠D = ∠B = 100° ... opposite angles.

Answer 2

GIVEN :-

★ Two adjacent anglies of a parallelogramm are in the

Two adjacent anglies of a parallelogramm are in theratio 4:5.

TO FIND :-

★ Measure of all angles .

SOLUTION :-

★ Let's,

$\red{ \bold{∠ A \: AND \: ∠B \: are \: two \: adjacent \: angles}}$

★ BUT WE KNOW THAT SUM OF ADJACENT ANGLES OF A PARALLELOGRAM IS 180°

$\red\longmapsto \large \bold \red{∠A + ∠B = 180°}$

★ ADJACENT ANGLES OF A PARALLELOGRAM ARE IN THE RATIO 4 : 5

☯ LET'S THE RATIO BE MULTIPLE OF X

➡ ∠A + ∠B = 180°

$\large\longmapsto \bold{4x + 5x = 180 {}^{ °} }$

$\large\longmapsto \bold{9x = 180°}$

$\large\longmapsto \bold{x = \cancel\frac{180}{9} }$

$\large \longmapsto \bold {x = \red{20°}}$

______________________

$\large \bold {∠ a = 4x = 4 \times 20 = 80 {}^{o} }$

$\large \bold {∠b = 5x = 5 \times 20 = 100 {}^{o} }$

$\red{ \large \bold{also \: ∠b +∠c = 180 {}^{o}} \bold \blue { \ \: \: \: \: \: \: (since \: ∠b \: and \:∠c \: are \: adajacent \: angles. ) }}$

$\large \bold{100 {}^{o} + ∠c = 180 {}^{o} }$

$\large \bold{∠c = 180 {}^{o} - 100 {}^{o} = 80 {}^{o} }$

NOW :-

$\bold{★ \: ∠C + ∠D = 180° \: \: \: \: \: \: \: [ SINCE \: ∠C \: AND \: ∠D \: ARE \: ADJACENT \: ANGLES ]}$

$\longmapsto \bold{8 0{}^{o} + ∠d = 180 {}^{o} }$

$\longmapsto \bold{∠d = 18 0{}^{o} - 80 {}^{0} = 100 {}^{o} }$